6.5: Black-Scholes Equation

Solutions of the Black-Scholes equation define the value of a derivative, for example of a call or put option, which is based on an asset. An asset can be a stock or a derivative of it, for instance. In principle, there are infinitely many such products, for example n-th derivatives. The Black-Scholes equation for the value $V(S,t)$ of a derivative is

$$\begin{equation} \label{BS1} V_t+\frac{1}{2}\sigma^2 S^2V_{SS}+rSV_S-rV=0\ \ \mbox{in}\ \Omega, \end{equation}$$

where for a fixed $T$, $0<T<\infty$,

$$\Omega=\{(S,t)\in\mathbb{R}^2:\ 0<S<\infty,\ 0<t<T\}, \]

and $\sigma$, $r$ are positive constants. More precisely, $\sigma$ is the volatility of the underlying asset $S$, $r$ is the guaranteed interest rate of a risk-free investment.

If $S(t)$ is the value of an asset at time $t$, then $V(S(t),t)$ is the value of the derivative at time $t$, where $V(S,t)$ is the solution of an appropriate initial-boundary value problem for the Black-Scholes equation, see below.

The Black-Scholes equation follows from Ito's Lemma under some assumptions on the random function associated to $S(t)$, see [26], for instance.

Call option

Here is $V(S,t):=C(S,t)$, where $C(S,t)$ is the value of the (European) call option. In this case we have following side conditions to ($\ref{BS1}$):

$$\begin{eqnarray} \label{BSC1} C(S,T)&=&\max\{S-E,0\}\\ \label{BSC2} C(0,t)&=&0\\ \label{BSC3} C(S,t)&=&S+o(S)\ \mbox{as}\ S\to\infty, \ \mbox{uniformly in}\ t, \end{eqnarray}$$

where $E$ and $T$ are positive constants, $E$ is the exercise price and $T$ the expiry.

Side condition ($\ref{BSC1}$) means that the value of the option has no value at time $T$ if $S(T)\le E$, condition ($\ref{BSC2}$) says that it makes no sense to buy assets if the value of the asset is zero, condition ($\ref{BSC3}$) means that we buy assets if its value becomes large, see Figure 6.5.1, where the side conditions are indicated.

Figure 6.5.1: Side conditions for a call option

Theorem 6.4 (Black-Scholes formula for European call options).
The solution $C(S,t)$, $0\le S<\infty$, $0\le t\le T$, of the initial-boundary value problem ($\ref{BS1}$)-($\ref{BSC3}$) is explicitly known and is given by

$$C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2),$$

where
$$\begin{eqnarray*} N(x)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\ e^{-y^2/2}\ dy,\\ d_1&=&\frac{\ln(S/E)+(r+\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}},\\ d_2&=&\frac{\ln(S/E)+(r-\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}. \end{eqnarray*}$$

Proof. Substitutions

$$S=Ee^x,\ \ t=T-\frac{\tau}{\sigma^2/2},\ \ C=Ev(x,\tau)\]

change equation ($\ref{BS1}$) to

$$\begin{equation} \label{BS2} v_\tau=v_{xx}+(k-1)v_x-kv, \end{equation}$$

where

$$k=\frac{r}{\sigma^2/2}.\]

Initial condition ($\ref{BS2}$) implies

$$\begin{equation} \label{BS3} v(x,0)=\max\{e^x-1,0\}. \end{equation}$$

For a solution of ($\ref{BS2}$) we make the ansatz

$$v=e^{\alpha x+\beta \tau}u(x,\tau),\]

where $\alpha$ and $\beta$ are constants which we will determine as follows. Inserting the ansatz into differential equation ($\ref{BS2}$), we get

$$\beta u+u_\tau=\alpha^2u+2\alpha u_x+u_{xx}+(k-1)(\alpha u+u_x)-ku.\]

Set $\beta=\alpha^2+(k-1)\alpha-k$ and choose $\alpha$ such that $0=2\alpha+(k-1)$, then $u_\tau=u_{xx}$. Thus

$$\begin{equation} \label{BS4} v(x,\tau)=e^{-(k-1)x/2-(k+1)^2\tau/4}u(x,\tau), \end{equation}$$

where $u(x,\tau)$ is a solution of the initial value problem

$$\begin{eqnarray*} u_\tau&=&u_{xx},\ \ -\infty<x<\infty,\ \tau>0\\ u(x,0)&=&u_0(x), \end{eqnarray*}$$

with

$$u_0(x)=\max\left\{e^{(k+1)x/2}-e^{(k-1)x/2},0\right\}.\]

A solution of this initial value problem is given by Poisson's formula

$$u(x,\tau)=\frac{1}{2\sqrt{\pi \tau}}\int_{-\infty}^{+\infty}\ u_0(s)e^{-(x-s)^2/(4\tau)}\ ds.\]

Changing variable by $q=(s-x)/(\sqrt{2\tau})$, we get
$$\begin{eqnarray*} u(x,\tau)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\ u_0(q\sqrt{2\tau}+x)e^{-q^2/2}\ dq\\ &=&I_1-I_2, \end{eqnarray*}$$
where
$$\begin{eqnarray*} I_1&=&\frac{1}{\sqrt{2\pi}}\int_{-x/(\sqrt{2\tau)}}^{\infty}\ e^{(k+1)(x+q\sqrt{2\tau})}e^{-q^2/2}\ dq\\ I_2&=&\frac{1}{\sqrt{2\pi}}\int_{-x/(\sqrt{2\tau)}}^{\infty}\ e^{(k-1)(x+q\sqrt{2\tau})}e^{-q^2/2}\ dq. \end{eqnarray*}$$
An elementary calculation shows that
$$\begin{eqnarray*} I_1&=&e^{(k+1)x/2+(k+1)^2\tau/4}N(d_1)\\ I_2&=&e^{(k-1)x/2+(k-1)^2\tau/4}N(d_2), \end{eqnarray*}$$
where
$$\begin{eqnarray*} d_1&=&\frac{x}{\sqrt{2\tau}}+\frac{1}{2}(k+1)\sqrt{2\tau}\\ d_2&=&\frac{x}{\sqrt{2\tau}}+\frac{1}{2}(k-1)\sqrt{2\tau}\\ N(d_i)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_i}\ e^{-s^2/2}\ ds,\ \ i=1,\ 2. \end{eqnarray*}$$

Combining the formula for $u(x,\tau)$, definition ($\ref{BS4}$) of $v(x,\tau)$ and the previous settings $x=\ln(S/E)$, $\tau=\sigma^2(T-t)/2$ and $C=Ev(x,\tau)$, we get finally the formula of Theorem 6.4.

In general, the solution $u$ of the initial value problem for the heat equation is not uniquely defined, see for example [10], pp. 206.

Uniqueness. The uniqueness follows from the growth assumption ($\ref{BSC3}$). Assume there are two solutions of ($\ref{BS1}$), ($\ref{BSC1}$)-($\ref{BSC3}$), then the difference $W(S,t)$ satisfies the differential equation ($\ref{BS1}$) and the side conditions

$$W(S,T)=0,\ W(0,t)=0,\ W(S,t)=O(S)\ \mbox{as}\ S\to\infty\]

uniformly in $0\le t\le T$.

From a maximum principle consideration, see an exercise, it follows that $|W(S,t)|\le cS$ on $S\ge 0$, $0\le t\le T$. The constant $c$ is independent of $S$ and $t$. From the definition of $u$ we see that
$$u(x,\tau)=\frac{1}{E}e^{-\alpha x-\beta\tau} W(S,t),$$
where $S=Ee^x$, $t=T-2\tau/(\sigma^2)$. Thus we have the growth property
$$\begin{equation} \label{wachs1} |u(x,\tau)|\le Me^{a|x|},\ \ x\in\mathbb{R}^1, \end{equation}$$
with positive constants $M$ and $a$. Then the solution of $u_\tau=u_{xx}$, in $-\infty<x<\infty$,
$0\le\tau\le\sigma^2 T/2$, with the initial condition $u(x,0)=0$ is uniquely defined in the class of functions satisfying the growth condition ($\ref{wachs1}$), see Proposition 6.2 of this chapter.
That is, $u(x,\tau)\equiv 0$.

$\Box$

Put option

Here is $V(S,t):=P(S,t)$, where $P(S,t)$ is the value of the (European) put option. In this case we have following side conditions to ($\ref{BS1}$):
$$\begin{eqnarray} \label{BSP1} P(S,T)&=&\max\{E-S,0\}\\ \label{BSP2} P(0,t)&=&Ee^{-r(T-t)}\\ \label{BSP3} P(S,t)&=&o(S)\ \mbox{as}\ S\to\infty,\ \mbox{uniformly in}\ 0\le t\le T. \end{eqnarray}$$
Here $E$ is the exercise price and $T$ the expiry.

Side condition ($\ref{BSP1}$) means that the value of the option has no value at time $T$ if $S(T)\ge E$, condition ($\ref{BSP2}$) says that it makes no sense to sell assets if the value of the asset is zero, condition ($\ref{BSP3}$) means that it makes no sense to sell assets if its value becomes large.

Theorem 6.5 (Black-Scholes formula for European put options).

The solution $P(S,t)$, $0<S<\infty$, $t<T$ of the initial-boundary value problem ($\ref{BS1}$), ($\ref{BSP1}$)-($\ref{BSP3}$) is explicitly known and is given by

$$P(S,t)=Ee^{-r(T-t)}N(-d_2)-SN(-d_1)$$
where $N(x)$, $d_1$, $d_2$ are the same as in Theorem 6.4.

Proof. The formula for the put option follows by the same calculations as in the case of a call option or from the put-call parity

$$C(S,t)-P(S,t)=S-Ee^{-r(T-t)}\]

and from

$$N(x)+N(-x)=1.\]

Concerning the put-call parity see an exercise. See also [26], pp. 40, for a heuristic argument which leads to the formula for the put-call parity.

$\Box$

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.